Page 2 of 2 FirstFirst 12
Results 16 to 23 of 23

Math Help - Range and inverse of bijective functions

  1. #16
    Member
    Joined
    Sep 2009
    Posts
    145
    Quote Originally Posted by Raoh View Post
    Yes the same steps
    f(x)=(x+1)^2\Rightarrow y=(x+1)^2\Leftrightarrow \sqrt{y}=x+1\Leftrightarrow x=\sqrt{y}-1.
    therefore f^{-1}(y)=\sqrt{y}-1.
    I'm trying

    Range = (-2, inf)?
    Last edited by mr fantastic; November 16th 2009 at 06:51 PM. Reason: Merged posts
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Member rowe's Avatar
    Joined
    Jul 2009
    Posts
    89
    Quote Originally Posted by Hellbent View Post
    Range = (-2, inf)?
    If f: A \to B, then f^{-1}: B \to A, provided the inverse exists. Look at what you defined your domain, or A, to be in the initial problem.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Member
    Joined
    Sep 2009
    Posts
    145
    Quote Originally Posted by rowe View Post
    If f: A \to B, then f^{-1}: B \to A, provided the inverse exists. Look at what you defined your domain, or A, to be in the initial problem.
    The domain is any number that's ≤ -2.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Member rowe's Avatar
    Joined
    Jul 2009
    Posts
    89
    Quote Originally Posted by Hellbent View Post
    The domain is any number that's ≤ -2.
    Right, so the range of the inverse cannot be (-2, \infty).
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Member
    Joined
    Sep 2009
    Posts
    145
    Quote Originally Posted by rowe View Post
    Right, so the range of the inverse cannot be (-2, \infty).
    I know that the range is the set of output numbers, but I'm not understanding it here.
    (1, inf)?
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Member rowe's Avatar
    Joined
    Jul 2009
    Posts
    89
    Quote Originally Posted by Hellbent View Post
    I know that the range is the set of output numbers, but I'm not understanding it here.
    (1, inf)?
    In your above problem, you have given:

    f : x |→ (x + 1) (x ∈ R, x ≤ -2)?

    So, we know the domain of f to be (-\infty, -2]. We know also know that if f: A \to B, then f^{-1}:B \to A, and we can therefore conclude that the range of f^{-1} is (-\infty, -2].
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    THE DOMAIN OF f IS THE RANGE OF f^{-1} AND VICE VERSA.
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Member
    Joined
    Sep 2009
    Posts
    145
    Thanks, rowe
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Bijective functions
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: April 1st 2011, 02:14 PM
  2. domain and range of inverse trig functions
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: October 14th 2009, 04:47 AM
  3. Proving : existance of the inverse/ bijective
    Posted in the Calculus Forum
    Replies: 8
    Last Post: May 13th 2009, 08:28 PM
  4. I need help on inverse of bijective function.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 28th 2008, 05:38 PM
  5. Proving bijective functions
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: November 29th 2007, 12:31 PM

Search Tags


/mathhelpforum @mathhelpforum